Analycity of Gamma function

I am stuck with something that may be simple, but I just cannot find the answer.

We define, for $$z$$ a complex number of real part strictly positive, $$\Gamma(z) = \int_{0}^{\infty} t^{z - 1}exp(-t) dt$$. How do we show that $$\Gamma$$ is an analytic function on $$\{ z \in \mathbb{C}, \Re(z) > 0 \}$$ ?

I started by rewritting the expression of $$\Gamma$$ this way:

$$\Gamma(z) = \int_{- \infty}^{\infty} exp(zu)exp(-e^{u}) du$$

Then, if $$a$$ is a complex number with $$\Re(a) > 0$$, we have:

$$\Gamma(z) = \int_{- \infty}^{\infty} exp((z - a)u)exp(au)exp(-e^{u}) du = \int_{- \infty}^{\infty} \sum \limits_{n = 0}^{+\infty} \frac{(z-a)^{n}}{n!}u^{n}exp(-e^{u}) du$$.

However after that, I do not see how I could exchange $$\sum$$ and $$\int$$ ...

It is analytical, because it is differentiable as a function of a complex variable $$z$$. The integral corresponding to the derivative converges uniformly, so you can differentiate with respect to the parameter.
• @JackEight Well, if you're trying to prove that $\Gamma$ is just a real analytic function, use the same approach: differentiate the integral w.r.t. the parameter and prove that all obtained integrals for all derivatives converge uniformly. But if you want to prove the statement for $\{z \in \mathbb{C}: \Re (z) > 0\}$, you should know some basic complex analysis. Feb 4, 2021 at 20:30